Is $\prod_{m=1}^{\infty}\frac{\Gamma(2mx+x+1)}{(2mx)^{2x+1}\Gamma(2mx-x)}=\frac{2x^{x+\frac 12}}{\pi^x\ \Gamma(x+1)}$ also for $x\notin\mathbb Z^+$? By calling upon the roots of unity, it can be proved that for $n\in\mathbb Z^+$, $$\prod_{k=1}^{n-1} \sin\frac{k\pi}{2n}= \frac{\sqrt n}{2^{n-1}} $$ (see, for example, this.)
Using the infinite product representation for $\sin x$, the LHS becomes $$\prod_{k=1}^n \frac{k\pi}{2n}\prod_{m=1}^{\infty}\left( 1-\frac{k^2}{4m^2n^2}\right)\\ = n!\left(\frac{\pi}{2n} \right)^n \prod_{m=1}^{\infty}\prod_{k=1}^n \frac{(2mn-k)(2mn+k)}{4m^2n^2}\\ = n!\left(\frac{\pi}{2n}\right)^n \prod_{m=1}^{\infty}\frac{\Gamma(2mn+n+1)}{(2mn)^{2n+1}\Gamma(2mn-n)} $$ and therefore
$$  \prod_{m=1}^{\infty}\frac{\Gamma(2mx+x+1)}{(2mx)^{2x+1}\Gamma(2mx-x)}= \frac{2x^{x+\frac 12}}{\pi^x \ \Gamma(x+1)} $$ for $x\in\mathbb Z^+$. Naturally, I wondered whether this result also holds for other $x\notin\mathbb Z^+$, and indeed on checking I saw that this is true for $x=\frac 12, \frac 32, \frac 13$ or even $x=\pi$, so I suspect that it holds atleast for all $x\gt 0$. How can this be proven/disproven?
 A: Suppose that $|\arg x|<\pi$. Then, by http://dlmf.nist.gov/5.11.E13,
$$
\frac{{\Gamma (2mx + x + 1)}}{{(2mx)^{2x + 1} \Gamma (2mx - x)}} = 1 - \frac{{(2x + 1)(x + 1)}}{{24x}}\frac{1}{{m^2 }}+\mathcal{O}_x\!\left(\frac{1}{m^3}\right)
$$
as $m\to +\infty$. Thus, the product on the LHS converges uniformly on compact subsets of $|\arg x|<\pi$ and defines an analytic function on $|\arg x|<\pi$.
Now assume that $|\arg x|\leq\frac{\pi}{2}$. By Stirling's formula
\begin{align*}
& \frac{{\Gamma (2mx + x + 1)}}{{(2mx)^{2x + 1} \Gamma (2mx - x)}} \sim \frac{{(2m + 1)^{2mx + x + 1/2} }}{{(2m)^{2x + 1} (2m - 1)^{2mx - x - 1/2} e^{2x} }} \\ &= \left( {1 - \frac{1}{{2m}}} \right)^{2x + 1} \left( {1 + \frac{2}{{2m - 1}}} \right)^{x + 1/2} \left( {\left( {1 + \frac{2}{{2m - 1}}} \right)^m \frac{1}{e}} \right)^{2x} 
\end{align*}
as $x\to \infty$, uniformly in $m\geq 1$. The relative error in this approximation is
$$
\frac{{1 + \frac{1}{{12x(2m + 1)}} + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|^2 m^2 }}} \right)}}{{1 + \frac{1}{{12x(2m - 1)}} + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|^2 m^2 }}} \right)}} = 1 + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|m^2 }}} \right).
$$
Thus
\begin{align*}
& \left| {\prod\limits_{m = 1}^\infty  {\frac{{\Gamma (2mx + x + 1)}}{{(2mx)^{2x + 1} \Gamma (2mx - x)}}} } \right| \\ & \ll \prod\limits_{m = 1}^\infty  {\left( {1 - \frac{1}{{2m}}} \right)^{2\Re x + 1} \left( {1 + \frac{2}{{2m - 1}}} \right)^{\Re x + 1/2} \left( {\left( {1 + \frac{2}{{2m - 1}}} \right)^m \frac{1}{e}} \right)^{2\Re x} }
\\ & \le \prod\limits_{m = 1}^\infty  {\exp \left( { - \frac{{2\Re x + 1}}{{2m}}} \right)\exp \left( {\frac{{2\Re x + 1}}{{2m - 1}}} \right)\exp \left( {\frac{{2\Re x}}{{10m^2 }}} \right)} \\ &=
\prod\limits_{m = 1}^\infty  {\exp \left( {\frac{{2\Re x + 1}}{{2m(2m - 1)}}} \right)\exp \left( {\frac{{2\Re x}}{{10m^2 }}} \right)} \\ & = 2\exp \left( {2\Re x\left( {\log 2 + \frac{{\pi ^2 }}{{60}}} \right)} \right) = \mathcal{O}(e^{2\left| x \right|} ).
\end{align*}
The RHS of your expression, which is analytic for $|\arg x|<\pi$, is
$$
\frac{{2x^{x + \frac{1}{2}} }}{{\pi ^x \Gamma (x + 1)}} \sim \sqrt {\frac{2}{{\pi x}}} \left( {\frac{e}{\pi }} \right)^x  = \mathcal{O}(e^{\left| x \right|} )
$$
for $|\arg x|\leq\frac{\pi}{2}$, by Stirling's formula. Thus, the LHS minus the RHS is $\mathcal{O}(e^{2\left| x \right|} )$ for $|\arg x|\leq\frac{\pi}{2}$ and is identically zero on the positive integers. By a theorem of Carlson (see below), the difference is identically zero on $|\arg x|\leq\frac{\pi}{2}$ too. By analytic continuation, the difference is zero on the whole of $|\arg x|<\pi$.
In summary,
$$  \prod_{m=1}^{\infty}\frac{\Gamma(2mx+x+1)}{(2mx)^{2x+1}\Gamma(2mx-x)}= \frac{2x^{x+\frac 12}}{\pi^x \ \Gamma(x+1)} $$
for $|\arg x|<\pi$.
Carlson's theorem: If $f(z)$ is holomorphic in the sector $|\arg z|\leq \alpha$ with $\alpha\geq \frac{\pi}{2}$, $|f(z)|=\mathcal{O}(e^{c|z|})$ with some $c < \pi$ in this sector, and if $f(n) = 0$ when $n = 1, 2, 3,\ldots$, then $f(z) \equiv  0$.
For more on this theorem, see G. H. Hardy, On two theorems of F. Carlson and S. Wigert, Acta Math. 42 (1920), pp. 327–339.
